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Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP Show that is an equivalence relation.

User Cerebrou
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Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that
A=J^(-1)AJ. Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that
A=P^(-1)BP. In this equality we can perform a right multiplication by
P^(-1) and obtain
AP^(-1) =P^(-1)B. Then, in the obtained equality we perform a left multiplication by P and get
PAP^(-1) =B. If we write
Q=P^(-1) and
Q^(-1) = P we have
B = Q^(-1)AQ. Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have
A=P^(-1)BP and from B↔C we have
B=Q^(-1)CQ. Now, if we substitute the last equality into the first one we get


A=P^(-1)Q^(-1)CQP = (P^(-1)Q^(-1))C(QP).

Recall that if P and Q are invertible, then QP is invertible and
(QP)^(-1)=P^(-1)Q^(-1). So, if we denote R=QP we obtained that


A=R^(-1)CR. Hence, A↔C.

Therefore, the relation is an equivalence relation.

User Paul Beusterien
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